# UVA-10608 Friends 【并查集】

2019/7/21 18:44:15 人评论 次浏览 分类：学习教程

There is a town with N citizens. It is known that some pairs of people are friends. According to the
famous saying that “The friends of my friends are my friends, too” it follows that if A and B are friends
and B and C are friends then A and C are friends, too.
Your task is to count how many people there are in the largest group of friends.

Input
Input consists of several datasets. The first line of the input consists of a line with the number of test
cases to follow.
The first line of each dataset contains tho numbers N and M, where N is the number of town’s
citizens (1 ≤ N ≤ 30000) and M is the number of pairs of people (0 ≤ M ≤ 500000), which are known
to be friends. Each of the following M lines consists of two integers A and B (1 ≤ A ≤ N, 1 ≤ B ≤ N,
A ̸= B) which describe that A and B are friends. There could be repetitions among the given pairs.

Output
The output for each test case should contain (on a line by itself) one number denoting how many people
there are in the largest group of friends on a line by itself.

Sample Input
2
3 2
1 2
2 3
10 12
1 2
3 1
3 4
5 4
3 5
4 6
5 2
2 1
7 1
1 2
9 10
8 9

Sample Output
3
7

`#include   using namespace std;    const int maxn = 10001;  int N, M;  int ans;  int p[maxn];  int cnt[maxn];    int ffind(int x)  {      return p[x] == x? x : p[x] = ffind(p[x]);  }    void init(int n)  {      ans = 1;      for(int i = 1; i <= n ; i++)          p[i] = i, cnt[i] = 1;    }    void uunion(int u, int v)  {      int a = ffind(u);      int b = ffind(v);      if(a != b)          p[b] = a, cnt[a] += cnt[b];      ans = max(cnt[a],ans);  }      int main()  {  #ifndef ONLINE_JUDGE      freopen("in.ini", "r",stdin);  #endif // ONLINE_JUDGE      int T;      cin >> T;      while(T--)      {          cin >> N >> M;          init(N);          int a,b;          for(int i = 0; i < M; i++)          {              cin >> a >> b;              uunion(a,b);          }            cout << ans << endl;      }      return 0;  }`

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